Question

Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method?
$x-3y-7=0,3x-3y-15=0$

• A. No solutions
• B. Unique solutions, $x=4,y=-1$
• C. Infinitely many solutions
• D. None of these

Answer 1

The correct option is B Unique solutions, $x=4,y=-1$

Given pair of equations are:
$x-3y-7=0$ ......(1)
$3x-3y-15=0$ ......(2)
$a_1=1,b_1=-3,c_1=-7$
$a_2=3,b_2=-3,c_2=-15$
$\Rightarrow \frac{a_1}{a_2}=\frac{1}{3},\frac{b_1}{b_2}=\frac{-3}{-3}$
$\Rightarrow \frac{a_1}{a_2}=\frac{1}{3},\frac{b_1}{b_2}=1$
It is clear that,
$\Rightarrow \frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
Hence, there will be unique solution for the given pair of linear equations.
From Cross multiplication Method:
$\frac{x}{b_1{c}_2-b_2{c}_1}=\frac{y}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$
$\Rightarrow \frac{x}{(-3\times -15)-(-3\times -7)}=\frac{y}{(-7\times 3)-(-15\times 1)}=\frac{1}{1(-3)-3(-3)}$
$\Rightarrow \frac{x}{45-21}=\frac{y}{-21+15}=\frac{1}{-3+9}$
$\Rightarrow \frac{x}{24}=\frac{y}{-6}=\frac{1}{6}$
$\Rightarrow \frac{x}{24}=\frac{1}{6}$
$\Rightarrow 6x=24$

$\Rightarrow x=4$ ....(3)
$\Rightarrow \frac{y}{-6}=\frac{1}{6}$
$\Rightarrow 6y=-6$

$\Rightarrow y=-1$ ....(4)
Hence, $x=4,y=-1$ .... From (3) and (4)